Pressure and Pascal's Law — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Pressure and Pascal's Law MCQs with step-by-step solutions (5 questions). Part of Mechanical Properties of Fluids. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Pressure and Pascal's Law · medium · numerical
A barometer is constructed using a liquid (density $= 760$ kg/m$^3$). What would be the height of the liquid column, when a mercury barometer reads 76 cm? (Density of mercury $= 13600$ kg/m$^3$)
A. 1.36 m
B. 13.6 m ✓ Correct
C. 136 m
D. 0.76 m
Solution: $\rho_l = 760$ kg/m$^3$, $\rho_m = 13600$ kg/m$^3$, $h_m = 76$ cm $= 0.76$ m. Equating pressures $p_{liquid} = p_{mercury} \Rightarrow h_l\rho_l g = h_m\rho_m g \Rightarrow h_l = \dfrac{h_m\rho_m}{\rho_l} = \dfrac{0.76\times13600}{760} = 13.6$ m.
Q2 — Pressure and Pascal's Law · medium · numerical
In a $U$-tube as shown in a figure, water and oil are in the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are 15 cm and 20 cm respectively. The density of the oil is [take $\rho_{water} = 1000$ kg/m$^3$]
A. 1200 kg/m$^3$
B. 750 kg/m$^3$ ✓ Correct
C. 1000 kg/m$^3$
D. 1333 kg/m$^3$
Solution: Pressure due to water column of height 15 cm $=$ pressure due to oil column of height 20 cm. $h_w\rho_w g = h_0\rho_0 g \Rightarrow 15\rho_w = 20\rho_0 \Rightarrow \rho_0 = \dfrac{15}{20}\rho_w = \dfrac{15}{20}\times1000 = 750$ kg/m$^3$.
Q3 — Pressure and Pascal's Law · medium · numerical
A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is
A. 650 kg/m$^3$
B. 425 kg/m$^3$
C. 800 kg/m$^3$
D. 928 kg/m$^3$ ✓ Correct
Solution: Both ends are open, so pressures at the same horizontal level are equal, $p_1 = p_2 \Rightarrow h_{oil}\cdot S_{oil}\cdot g = h_{water}\cdot S_{water}\cdot g$. From the figure $S_{oil} = \dfrac{(65+65)\times1000}{(65+65+10)} = 928$ kg/m$^3$.
Q4 — Pressure and Pascal's Law · medium · numerical
Two non-mixing liquids of densities $\rho$ and $n\rho$ $(n > 1)$ are put in a container. The height of each liquid is $h$. A solid cylinder of length $L$ and density $d$ is put in this container. The cylinder floats with its axis vertical and length $pL$ $(p < 1)$ in the denser liquid. The density $d$ is equal to
A. $\{2 + (n+1)p\}\rho$
B. $\{2 + (n-1)p\}\rho$
C. $\{1 + (n-1)p\}\rho$ ✓ Correct
D. $\{1 + (n+1)p\}\rho$
Solution: By Archimedes principle, weight of cylinder $=$ (upthrust)$_1 +$ (upthrust)$_2$, i.e. $ALdg = (1-p)LA\rho g + (pLA)n\rho g \Rightarrow d = (1-p)\rho + pn\rho = \rho - p\rho + np\rho = \rho + (n-1)p\rho = \rho[1 + (n-1)p]$.
Q5 — Pressure and Pascal's Law · medium · numerical
The approximate depth of an ocean is 2700 m. The compressibility of water is $45.4\times10^{-11}$ Pa$^{-1}$ and density of water is $10^3$ kg/m$^3$. What fractional compression of water will be obtained at the bottom of the ocean?
A. $0.8\times10^{-2}$
B. $1.0\times10^{-2}$
C. $1.2\times10^{-2}$ ✓ Correct
D. $1.4\times10^{-2}$
Solution: Pressure at the bottom of the ocean $p = \rho g d = 10^3\times10\times2700 = 27\times10^6$ Pa. Fractional compression $=$ compressibility $\times$ pressure $= 45.4\times10^{-11}\times27\times10^6 = 1.2\times10^{-2}$.